ME-662 CONVECTIVE HEAT AND MASSTRANSFER

A. W. DateMechanical Engineering Department

Indian Institute of Technology, BombayMumbai - 400076

India

LECTURE-26 TURBULENCE MODELS-1

() March 9, 2011 1 / 1

LECTURE-26 TURBULENCE MODELS-1

1 Main Task2 Eddy Viscosity Models

1 General Mixing Length model2 High Ret One-Eqn model3 High Ret Two-Eqn model

() March 9, 2011 2 / 1

Main Tasks - L26( 120)

1 In multidimensional turbulent wall flows, even if inner-layeruniversality is exploited , RANS eqns must be solved in theouter layers through modeling

2 Thus, turbulent stresses u′i u′

j and heat fluxes ρ cp u′i T ′

must be modeled to recover lost information throughaveraging

3 This recovery must be carried out in a general way so thatthe model need not be changed from one flow situation toanother

4 Imparting absolute generality to turbulence models has,however, turned out to be quite a difficult task.

5 Thirdly, the model must be economical; that is, thecomputational expense ( and ease ) must not be very muchin excess of that which would be required for computation ofa laminar flow under the same situation.

() March 9, 2011 3 / 1

Two Main Approaches - L26( 220)

1 Relating u′i u′

j to the mean rate of strain Sij through aproperty called the turbulent- or eddy- viscosity µt . Thisapproach derives its inspiration from the Stokes’sstress-strain relations.

2 Recovering distribution of stresses from solution oftransport equations for u′

i u′j in which convection and

diffusion of this quantity is principally balanced by rates ofits production and dissipation.

In this lecture, we shall consider the most popularEddy-viscosity models

() March 9, 2011 4 / 1

The main idea - L26( 320)

1 The notion of µt introduced by Boussinesque can begeneralised to read as:

−ρ u′i u′

j = µt

[∂ui

∂xj+

∂uj

∂xi

]− 2

3ρ e δij

where, µt is a property of the flow; not that of a fluid . µt isisotropic although its magnitude may vary with the positionin the flow .

2 The term involving Kronecker delta δij simply ensures thatthe sum of the normal stresses ( i = j ) will equal 2 ρ e andthus, the definition of TKE is retrieved since the sum of thestrain rates ( ∂ui/∂xi = 0 ) is zero from requirement ofcontinuity.

3 Using this model, number of unknowns ( ui and p ) equalsnumber of RANS eqns when µt is modeled.

() March 9, 2011 5 / 1

Characterising µt - L26( 420)

1 From kinetic theory, laminar viscosity µ is written indimensionally correct form as

µ = ρ× lmfpl × umol

where lmfpl is mean free path length and umol is averagemolecular velocity.

2 Analogously, µt is modeled as

µt = ρ× L× |u′|

where L and |u′| are representative length and velocityscales of turbulent fluctuations.

3 The main task now is to represent L and |u′| universally.

() March 9, 2011 6 / 1

General Mixing Length Model - L26( 520)

1 Simplest model - |u′| ∝ lm × |mean velocity gradient|2 Hence,

µt = ρm l2m

{∂ui

∂xj(∂ui

∂xj+

∂uj

∂xi)

}0.5

summation,

where L = lm is called Prandtl’s mixing length

lm = κ× n × (1− e−ξ) ξ =1

26nν

√τw

ρm=

n+

26

lm = 0.2× κ× R , if n+ > 26 ,

where n is the normal distance from the nearest wall andκ ' 0.41, R is a characteristic dimension, and wall shearstress τw is evaluated from the product of µ and total velgradient ∂V/∂n|wall

() March 9, 2011 7 / 1

lm for 2D Boundary Layers - L26( 620)

1 For 2D BLs, µt = ρm l2m (∂u/∂y)

2 For inner and outer wall boundary layers, from lecture 24

lm = κ× y × (1− e−y+

A+ )

A+ =25

a[v+

w + b{

p+

1+c v+w

}]+ 1

3 For Free Shear Layers such as jets and wakes,

lm = β y1/2

where y1/2 is the half-width of the shear layer andβ = 0.225 ( a plane jet ), = 0.1875 ( a round jet ),= 0.40 ( a plane wake ).

() March 9, 2011 8 / 1

One Eqn Model - 1 - L26( 720)

1 The mixing length model predicts that −ρ u′i u′

j = 0 wherethe strain rate Sij = 0. In many situations this is not found.For example, in an annulus with outer wall rough and innerwall smooth, the plane of zero shear stress is closer to thesmooth wall than the plane of zero vel gr.

2 Hence, we take the fluctuating velocity scale |u′| =√

e .Then, µt = ρ× L×

√e where L = Integral Length Scale .

3 The distribution of TKE (e) is determined from

ρ

[∂e∂t

+ uj∂e∂xj

]= − ∂

∂xj

[u′

j (p′ + ρ

u′i u

′i

2)− µ

∂e∂xj

]

+ (−ρu′i u

′j )

∂ui

∂xj− µ (

∂u′i

∂xj)2

() March 9, 2011 9 / 1

Modeling TKE - L26( 820)

1 The turbulent diffusion term cannot be directly measuredbecause no probe can simultaneously measure pressureand velocity fluctuations

2 But, noting the redistributive character of this term, we mayassume gradient diffusion. Hence,

−u′j (p

′ + ρu′

i u′i

2) =

µt

σe

∂e∂xj

where, σe is a turbulent Prandtl number for TKE.3 Recall that when Ret is high, dissipation can be

represented in terms of large scale fluctuations. Hence,

µ (∂u′

i

∂xj)2 = CD

ρ e3/2

L= ρ ε

() March 9, 2011 10 / 1

Modeled TKE Eqn L26( 920)

Replacing −ρ u′i u′

j = µt Sij

ρ

[∂e∂t

+ uj∂e∂xj

]=

∂

∂xj

[(µ +

µt

σe)

∂e∂xj

]+ µt

[∂ui

∂xj+

∂uj

∂xi

]∂ui

∂xj− CD

ρ e3/2

L

where CD and σe are expected to be universal constantswhen Ret = L

√e/ν is high ( that is away from the wall and

beyond the transition layer ).

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Determination of CD and σe - L26( 1020)

1 Recall that in the fully turbulent inner layer, production 'dissipation ( lecture 24 ) and equilibrium conditions prevail.

2 In this layer, τt = µt ∂V/∂n. Hence

τt∂V∂n

= ρ√

e L (∂V∂n

)2 = CDρ e3/2

L

where V is vel. parallel to the wall and n is normal distance.3 Also, in this layer τt ' τw . Hence, (τw/ρ)/ e = u2

τ / e =√

CD

4 Recall that τw ' 0.3 ρ e. Hence, CD ' 0.095 σe is taken as 1.0 from numerical experiments in several

flow situations.6 Hence, the modeled TKE eqn can be solved. We must now

specify L.

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Specification of L - L26( 1120)

1 L is determined as follows. Consider

µt (∂V∂n

)2 = CDρ e3/2

L(equilibrium condition)

µt = ρ e0.5 L (definition)

2 Eliminating e from these two equations,µt = C−0.5

D ρ L2 (∂V/∂n)3 Comparing this equation with eqn with mixing length model

L = C0.25D lm = 0.5477 lm → lm = κ y

4 With these specifications of L, µt , CD and σe, TKE equationcan be solved along with the RANS momentum equations.In general flows, however, further refinements are needed.

() March 9, 2011 13 / 1

Two-Equation Model - L26( 1220)

1 For the more general case of turbulent flows involvingstrong and variable pressure gradients, flow recirculation,effects of swirl or buoyancy etc.,it is necessary to devisemeans for determining distribution of L in multidimensionalflows.

2 Dividing the energy spectrum e(k) by wave number k andintegrating from 0 to ∞, an equation for L in the spectralspace can indeed be derived since, L = 1

e

∫∞0

e(k)k dk .

However, the eqn is not tractable in the physical space.3 A length-scale equation, however, need not necessarily

have L itself as its dependent variable; any combination ofthe form Z = em Ln will suffice since e can be known fromsolution of modeled TKE equation

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Proposals for Z = em Ln Eqn - L26( 1320)

ρ

[∂Z∂t

+ uj∂Z∂xj

]=

∂

∂xj

[(µ +

µt

σZ)

∂Z∂xj

]+ C1

Ze

P − C2 Zρ√

eL

+ S (Z )

where, P = −ρu′i u

′j ∂ui/∂xj and C1, C2 are constants when

Ret = e0.5L/ν is high.Proposals for Z = em Ln

m n Remark3/2 -1 Dissipation Rate ( ε Eqn )1 1 ( eL Eqn )1 - 2 Vorticity Fluctuation ( e / L2 Eqn )

1/2 1 Turbulent viscosity ( e0.5 L Eqn )Computational experience suggests that for Z = (e3/2/ L) ∝ ε,S ( Z ) = 0 and σZ = const . Hence, Dissipation eqn preferred .

() March 9, 2011 15 / 1

Dissipation eqn - 1 - L26( 1420)

At high Ret , local isotropy prevails. Hence an eqn forε = ν (∂u′

i /∂xj)2 can be derived by differentiating N-S Eqn for u′

iw.r.t. xj and then multiplying by 2 ν (∂u

′

i /∂xj). Time averaginggives exact ε Eqn .[

∂ε

∂t+ uk

∂ε

∂xk

]= −ν

∂

∂xk

[u′

k (∂u′

i

∂xj)2 +

1ρ

∂p′

∂xj

∂u′k

∂xj− ∂ε

∂xk

]

− 2 ν

[∂u′

i

∂xj

∂u′k

∂xj+

∂u′j

∂xi

∂u′j

∂xk

]∂ui

∂xk

− 2 ν∂u′

i

∂xk

∂u′i

∂xj

∂u′k

∂xj− 2

[ν

∂2 u′i

∂xk ∂xj

]2

() March 9, 2011 16 / 1

Dissipation eqn - 2 - L26( 1520)

Complex correlations can only be discerned from DNS.

Diffusion = −ν∂

∂xk(u′

k (∂u′

i

∂xj)2 +

1ρ

∂p′

∂xj

∂u′k

∂xj)

=∂

∂xk

{C3

e2

ε

∂ε

∂xk

}→ u′

j u′k ∝ e (assumption)

Generation = − 2 ν (∂u′

i

∂xj

∂u′k

∂xj+

∂u′j

∂xi

∂u′j

∂xk)

∂ui

∂xk

= C1 u′i u

′j

ε

e∂ui

∂xj

last 2 terms = − 2 ν∂u′

i

∂xk

∂u′i

∂xj

∂u′k

∂xj− 2 (ν

∂2 u′i

∂xk ∂xj)2

= C2ε2

e(Relevant in inertial subrange)

() March 9, 2011 17 / 1

Dissipation eqn - 3 - L26( 1620)

Modeled dissipation equation[∂ε

∂t+ uk

∂ε

∂xk

]=

∂

∂xk

{(ν + C3

e2

ε)

∂ε

∂xk

}− C1 u′

i u′k

ε

e∂ui

∂xk− C2

ε2

e

But µt = ρ√

e L = CDρ e2

εHence,

ρ

[∂ε

∂t+ uk

∂ε

∂xk

]=

∂

∂xk

{(µ +

µt

σε

)∂ε

∂xk

}+ C1 (−ρ u′

i u′k) (

ε

e)

∂ui

∂xk− C2

ρ ε2

e

where −ρ u′i u

′k = µt [∂ui/∂xk + ∂uk/∂xi ] and

σε ≡ turbulent Prandtl number for dissipation rate. It absorbsconstants CD and C3.

() March 9, 2011 18 / 1

High Ret e-ε Model Eqns - L26( 1720)

ρD eDt

=∂

∂xj

[(µ +

µt

σe)

∂e∂xj

]+ µt

[∂ui

∂xj+

∂uj

∂xi

]∂ui

∂xj− ρ ε

ρD ε

Dt=

∂

∂xj

[(µ +

µt

σε

)∂ε

∂xj

]+

ε

e

{C1 µt

[∂ui

∂xj+

∂uj

∂xi

]∂ui

∂xj− C2 ρ ε

}Boundary layer Forms

ρ

[∂e∂t

+ u∂e∂x

+ v∂e∂y

]=

∂

∂y

{(µ +

µt

σe)

∂e∂y

}+ µt (

∂u∂y

)2 − ρ ε

ρ

[∂ε

∂t+ u

∂ε

∂x+ v

∂ε

∂y

]=

∂

∂y

{(µ +

µt

σε

)∂ε

∂y

}+

ε

e

{C1 µt (

∂u∂y

)2 − C2 ρ ε

}() March 9, 2011 19 / 1

Determination of C1, C2 and σε - L26( 1820)

From the experimental data on decay of homogeneousturbulence behind a grid in a wind-tunnel, it is found that e ∝ t−n

where, for t → 0, 1 < n < 1.2. In this flow, both production anddiffusion are absent and v = 0. Hence

DeDt

=∂e∂t

+ u∂e∂x

= −ε andDε

Dt=

∂ε

∂t+ u

∂ε

∂x= −C2

ε2

e

Simultaneous solution gives C2 = (n + 1)/n. Therefore,taking n = 1.1 (say), C2 = 1.91.To determine C1, consider inner turbulent layer where conv = 0,νt >> ν and production νt (∂u/∂y)2 = ε dissipation. Hence εEqn will read as

∂

∂y

{µt

σε

∂ε

∂y

}=

ε2

e(C2 − C1)

() March 9, 2011 20 / 1

Continued . . . - L26( 1920)

In the inner layer, logarithmic law u+ = ln (E y+)/κ gives

∂u∂y

=uτ

κ y=

τw

ρ νt=

u2τ

νt→ νt = uτ κ y

Hence, since Prod = Diss

ε = νt (∂u∂y

)2 =u3

τ

κ y→ ∂ε

∂y= − u3

τ

κ y2

Therefore ε Eqn becomes

∂

∂y

{uτκ y

σε

× (− u3τ

κ y2)

}=

u4τ

σε y2=

u6τ

κ2 y2 e(C2 − C1)

Or,κ2

σε

= (C2 − C1)u2

τ

e→ C1 = C2 −

κ2

σε

√CD

= 1.44

where σε = 1.3 ( from num. comp. ), CD = 0.09 and κ = 0.41() March 9, 2011 21 / 1

Wall-Function BCs - L26( 2020)

1 Wall BCs are given at 1stgrid node in the inner turblayer

2 Then, τw = µeff (∂u/∂y)p =µeff (up/yp). Hence,µeff /yp = (ρ uτ κ)/ln (E y+

p )

where uτ = C0.25D e0.5

p .3 In e-Eqn Prod =

τw (∂u/∂y)p = µeff (up/yp)2

4 εp = y−1p

∫ yp

0 ε dy= y−1

p

∫ yp

0 (τw/ρ)(∂u/∂y) dy= (u2

τ up)/yp

= C0.75D e1.5

p ln (E y+p )/(κ yp)

P

YP YP+30 < < 100

WALL

Grid Lines

Hence, BCs are effected as

Sourcee = µeff (up/yp)2 − ρ εp

εp =C0.75

D e1.5p

κ yp

Gives computational economy.

() March 9, 2011 22 / 1